Q: Splitting vector into x and y components.











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I have struck a problem within the aspects of simple middle school geometry. I'll just cut to the case.



Visual exmaple of problem



We have a triangle of any size or form. We know angles A, B and C. We have a vector, v along one of the triangles sides. We know the magnitude of vector v. We have a smaller triangle representing the x and y components. We do not know the angles of the smaller triangle, other than it is a right angled triangle. Keep in mind that the original triangle can be of any size or form, so in the case of the picture, extending the vector triangle to get one of the angles is not an option. Is it possible to acquire the x and y components? If so, is there any general formula to achieve this?



Sorry for the horrible picture. Any help is highly appreciated.



Edit: I forgot to inform that we also know the x and y coordinates of the original triangles corners.










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  • It's not possible as the orientation of the original triangle is not known. If you can rotate the figure in such a way that vector is parallel to axis $x$ or $y$, you solved your problem.
    – Vasya
    Nov 15 at 15:41

















up vote
0
down vote

favorite












I have struck a problem within the aspects of simple middle school geometry. I'll just cut to the case.



Visual exmaple of problem



We have a triangle of any size or form. We know angles A, B and C. We have a vector, v along one of the triangles sides. We know the magnitude of vector v. We have a smaller triangle representing the x and y components. We do not know the angles of the smaller triangle, other than it is a right angled triangle. Keep in mind that the original triangle can be of any size or form, so in the case of the picture, extending the vector triangle to get one of the angles is not an option. Is it possible to acquire the x and y components? If so, is there any general formula to achieve this?



Sorry for the horrible picture. Any help is highly appreciated.



Edit: I forgot to inform that we also know the x and y coordinates of the original triangles corners.










share|cite|improve this question
























  • It's not possible as the orientation of the original triangle is not known. If you can rotate the figure in such a way that vector is parallel to axis $x$ or $y$, you solved your problem.
    – Vasya
    Nov 15 at 15:41















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have struck a problem within the aspects of simple middle school geometry. I'll just cut to the case.



Visual exmaple of problem



We have a triangle of any size or form. We know angles A, B and C. We have a vector, v along one of the triangles sides. We know the magnitude of vector v. We have a smaller triangle representing the x and y components. We do not know the angles of the smaller triangle, other than it is a right angled triangle. Keep in mind that the original triangle can be of any size or form, so in the case of the picture, extending the vector triangle to get one of the angles is not an option. Is it possible to acquire the x and y components? If so, is there any general formula to achieve this?



Sorry for the horrible picture. Any help is highly appreciated.



Edit: I forgot to inform that we also know the x and y coordinates of the original triangles corners.










share|cite|improve this question















I have struck a problem within the aspects of simple middle school geometry. I'll just cut to the case.



Visual exmaple of problem



We have a triangle of any size or form. We know angles A, B and C. We have a vector, v along one of the triangles sides. We know the magnitude of vector v. We have a smaller triangle representing the x and y components. We do not know the angles of the smaller triangle, other than it is a right angled triangle. Keep in mind that the original triangle can be of any size or form, so in the case of the picture, extending the vector triangle to get one of the angles is not an option. Is it possible to acquire the x and y components? If so, is there any general formula to achieve this?



Sorry for the horrible picture. Any help is highly appreciated.



Edit: I forgot to inform that we also know the x and y coordinates of the original triangles corners.







geometry vectors triangle






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edited Nov 15 at 15:48

























asked Nov 15 at 15:35









Lukas Knudsen

33




33












  • It's not possible as the orientation of the original triangle is not known. If you can rotate the figure in such a way that vector is parallel to axis $x$ or $y$, you solved your problem.
    – Vasya
    Nov 15 at 15:41




















  • It's not possible as the orientation of the original triangle is not known. If you can rotate the figure in such a way that vector is parallel to axis $x$ or $y$, you solved your problem.
    – Vasya
    Nov 15 at 15:41


















It's not possible as the orientation of the original triangle is not known. If you can rotate the figure in such a way that vector is parallel to axis $x$ or $y$, you solved your problem.
– Vasya
Nov 15 at 15:41






It's not possible as the orientation of the original triangle is not known. If you can rotate the figure in such a way that vector is parallel to axis $x$ or $y$, you solved your problem.
– Vasya
Nov 15 at 15:41












3 Answers
3






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oldest

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up vote
0
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accepted










OK, your edit makes all the difference in the world. Let $B(x_B,y_B)$ be the start of the vector and $A(x_A,y_A)$ is a vertex on the same side of the triangle where the vector is. Then you can easily find tangent of the vector $m=frac{Y_B-Y_A}{X_B-X_A}$ (beware of division by zero). Then $|x|=frac{1}{sqrt{1+m^2}}v$, $|y|=frac{|m|}{sqrt{1+m^2}}v$ where $v$ is the magnitude of vector $vec{v}$. You can determine whether $x$ and $y$ are positive or negative by finding components of vector $vec{BA}(x_A-x_B,y_A-y_B)$. They will have the same signs.






share|cite|improve this answer























  • Thank you so much for the answer. Is your labeling relative to the picture i posted? The math doesn't quite make sense to me. Sorry if it is too much to ask for.
    – Lukas Knudsen
    Nov 15 at 16:20










  • @LukasKnudsen: You have not labeled vertices in your picture, I guess you want vector to start at $C$ according to the $angle{C}$ so just replace $B$ with $C$ in my answer. The other vertex remains $A$. The math is basically finding sine and cosine based on tangent.
    – Vasya
    Nov 15 at 16:32










  • Thank you for the clarification. Just to make sure I understood it, here's an example. $A( 0, 0 ), B( 10, 5 ), v=5, m=tan^-1(frac{5 - 0}{10 - 0})=26.6, |x|=frac{5}{sqrt{1 + 26.6^2}}=0.188, |y|=frac{26.6 * 5}{sqrt{1 + 26.6^2}}=4.996$. The x and y components seem to be off. Is this correct?
    – Lukas Knudsen
    Nov 15 at 16:58












  • @LukasKnudsen: no, $m=frac{5-0}{10-0}=0.5$, $|x|approx 4.47$, $|y| approx 2.236$, $BA(-10,-5)$ so $x=-4.47, y=-2.236$
    – Vasya
    Nov 15 at 17:12












  • Ohhh, I see. $m$ is the gradient and I just need to multiply $|x|$ and $|y|$ by the components of the directional vector of $vec{BA}$. Thank you.
    – Lukas Knudsen
    Nov 15 at 17:18


















up vote
0
down vote













Not from the information provided. In particular, any rotation of the triangle shown would give different $x$ and $y$ values, but the same $A$, $B$, and $C$, so there is no possible function giving $x$ and $y$ in terms of $A$, $B$, and $C$ alone.






share|cite|improve this answer




























    up vote
    0
    down vote













    Since you know the coordinates of the vertices, you can ignore the angles. The coordinates of $vec v$ can be found as a proportion of the coordinates of the appropriate edge. For the illustrated case, $$vec v = {|vec v|over|A-C|}(A-C).$$






    share|cite|improve this answer





















      Your Answer





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      3 Answers
      3






      active

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      3 Answers
      3






      active

      oldest

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      active

      oldest

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      up vote
      0
      down vote



      accepted










      OK, your edit makes all the difference in the world. Let $B(x_B,y_B)$ be the start of the vector and $A(x_A,y_A)$ is a vertex on the same side of the triangle where the vector is. Then you can easily find tangent of the vector $m=frac{Y_B-Y_A}{X_B-X_A}$ (beware of division by zero). Then $|x|=frac{1}{sqrt{1+m^2}}v$, $|y|=frac{|m|}{sqrt{1+m^2}}v$ where $v$ is the magnitude of vector $vec{v}$. You can determine whether $x$ and $y$ are positive or negative by finding components of vector $vec{BA}(x_A-x_B,y_A-y_B)$. They will have the same signs.






      share|cite|improve this answer























      • Thank you so much for the answer. Is your labeling relative to the picture i posted? The math doesn't quite make sense to me. Sorry if it is too much to ask for.
        – Lukas Knudsen
        Nov 15 at 16:20










      • @LukasKnudsen: You have not labeled vertices in your picture, I guess you want vector to start at $C$ according to the $angle{C}$ so just replace $B$ with $C$ in my answer. The other vertex remains $A$. The math is basically finding sine and cosine based on tangent.
        – Vasya
        Nov 15 at 16:32










      • Thank you for the clarification. Just to make sure I understood it, here's an example. $A( 0, 0 ), B( 10, 5 ), v=5, m=tan^-1(frac{5 - 0}{10 - 0})=26.6, |x|=frac{5}{sqrt{1 + 26.6^2}}=0.188, |y|=frac{26.6 * 5}{sqrt{1 + 26.6^2}}=4.996$. The x and y components seem to be off. Is this correct?
        – Lukas Knudsen
        Nov 15 at 16:58












      • @LukasKnudsen: no, $m=frac{5-0}{10-0}=0.5$, $|x|approx 4.47$, $|y| approx 2.236$, $BA(-10,-5)$ so $x=-4.47, y=-2.236$
        – Vasya
        Nov 15 at 17:12












      • Ohhh, I see. $m$ is the gradient and I just need to multiply $|x|$ and $|y|$ by the components of the directional vector of $vec{BA}$. Thank you.
        – Lukas Knudsen
        Nov 15 at 17:18















      up vote
      0
      down vote



      accepted










      OK, your edit makes all the difference in the world. Let $B(x_B,y_B)$ be the start of the vector and $A(x_A,y_A)$ is a vertex on the same side of the triangle where the vector is. Then you can easily find tangent of the vector $m=frac{Y_B-Y_A}{X_B-X_A}$ (beware of division by zero). Then $|x|=frac{1}{sqrt{1+m^2}}v$, $|y|=frac{|m|}{sqrt{1+m^2}}v$ where $v$ is the magnitude of vector $vec{v}$. You can determine whether $x$ and $y$ are positive or negative by finding components of vector $vec{BA}(x_A-x_B,y_A-y_B)$. They will have the same signs.






      share|cite|improve this answer























      • Thank you so much for the answer. Is your labeling relative to the picture i posted? The math doesn't quite make sense to me. Sorry if it is too much to ask for.
        – Lukas Knudsen
        Nov 15 at 16:20










      • @LukasKnudsen: You have not labeled vertices in your picture, I guess you want vector to start at $C$ according to the $angle{C}$ so just replace $B$ with $C$ in my answer. The other vertex remains $A$. The math is basically finding sine and cosine based on tangent.
        – Vasya
        Nov 15 at 16:32










      • Thank you for the clarification. Just to make sure I understood it, here's an example. $A( 0, 0 ), B( 10, 5 ), v=5, m=tan^-1(frac{5 - 0}{10 - 0})=26.6, |x|=frac{5}{sqrt{1 + 26.6^2}}=0.188, |y|=frac{26.6 * 5}{sqrt{1 + 26.6^2}}=4.996$. The x and y components seem to be off. Is this correct?
        – Lukas Knudsen
        Nov 15 at 16:58












      • @LukasKnudsen: no, $m=frac{5-0}{10-0}=0.5$, $|x|approx 4.47$, $|y| approx 2.236$, $BA(-10,-5)$ so $x=-4.47, y=-2.236$
        – Vasya
        Nov 15 at 17:12












      • Ohhh, I see. $m$ is the gradient and I just need to multiply $|x|$ and $|y|$ by the components of the directional vector of $vec{BA}$. Thank you.
        – Lukas Knudsen
        Nov 15 at 17:18













      up vote
      0
      down vote



      accepted







      up vote
      0
      down vote



      accepted






      OK, your edit makes all the difference in the world. Let $B(x_B,y_B)$ be the start of the vector and $A(x_A,y_A)$ is a vertex on the same side of the triangle where the vector is. Then you can easily find tangent of the vector $m=frac{Y_B-Y_A}{X_B-X_A}$ (beware of division by zero). Then $|x|=frac{1}{sqrt{1+m^2}}v$, $|y|=frac{|m|}{sqrt{1+m^2}}v$ where $v$ is the magnitude of vector $vec{v}$. You can determine whether $x$ and $y$ are positive or negative by finding components of vector $vec{BA}(x_A-x_B,y_A-y_B)$. They will have the same signs.






      share|cite|improve this answer














      OK, your edit makes all the difference in the world. Let $B(x_B,y_B)$ be the start of the vector and $A(x_A,y_A)$ is a vertex on the same side of the triangle where the vector is. Then you can easily find tangent of the vector $m=frac{Y_B-Y_A}{X_B-X_A}$ (beware of division by zero). Then $|x|=frac{1}{sqrt{1+m^2}}v$, $|y|=frac{|m|}{sqrt{1+m^2}}v$ where $v$ is the magnitude of vector $vec{v}$. You can determine whether $x$ and $y$ are positive or negative by finding components of vector $vec{BA}(x_A-x_B,y_A-y_B)$. They will have the same signs.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 15 at 16:15

























      answered Nov 15 at 16:09









      Vasya

      3,2691515




      3,2691515












      • Thank you so much for the answer. Is your labeling relative to the picture i posted? The math doesn't quite make sense to me. Sorry if it is too much to ask for.
        – Lukas Knudsen
        Nov 15 at 16:20










      • @LukasKnudsen: You have not labeled vertices in your picture, I guess you want vector to start at $C$ according to the $angle{C}$ so just replace $B$ with $C$ in my answer. The other vertex remains $A$. The math is basically finding sine and cosine based on tangent.
        – Vasya
        Nov 15 at 16:32










      • Thank you for the clarification. Just to make sure I understood it, here's an example. $A( 0, 0 ), B( 10, 5 ), v=5, m=tan^-1(frac{5 - 0}{10 - 0})=26.6, |x|=frac{5}{sqrt{1 + 26.6^2}}=0.188, |y|=frac{26.6 * 5}{sqrt{1 + 26.6^2}}=4.996$. The x and y components seem to be off. Is this correct?
        – Lukas Knudsen
        Nov 15 at 16:58












      • @LukasKnudsen: no, $m=frac{5-0}{10-0}=0.5$, $|x|approx 4.47$, $|y| approx 2.236$, $BA(-10,-5)$ so $x=-4.47, y=-2.236$
        – Vasya
        Nov 15 at 17:12












      • Ohhh, I see. $m$ is the gradient and I just need to multiply $|x|$ and $|y|$ by the components of the directional vector of $vec{BA}$. Thank you.
        – Lukas Knudsen
        Nov 15 at 17:18


















      • Thank you so much for the answer. Is your labeling relative to the picture i posted? The math doesn't quite make sense to me. Sorry if it is too much to ask for.
        – Lukas Knudsen
        Nov 15 at 16:20










      • @LukasKnudsen: You have not labeled vertices in your picture, I guess you want vector to start at $C$ according to the $angle{C}$ so just replace $B$ with $C$ in my answer. The other vertex remains $A$. The math is basically finding sine and cosine based on tangent.
        – Vasya
        Nov 15 at 16:32










      • Thank you for the clarification. Just to make sure I understood it, here's an example. $A( 0, 0 ), B( 10, 5 ), v=5, m=tan^-1(frac{5 - 0}{10 - 0})=26.6, |x|=frac{5}{sqrt{1 + 26.6^2}}=0.188, |y|=frac{26.6 * 5}{sqrt{1 + 26.6^2}}=4.996$. The x and y components seem to be off. Is this correct?
        – Lukas Knudsen
        Nov 15 at 16:58












      • @LukasKnudsen: no, $m=frac{5-0}{10-0}=0.5$, $|x|approx 4.47$, $|y| approx 2.236$, $BA(-10,-5)$ so $x=-4.47, y=-2.236$
        – Vasya
        Nov 15 at 17:12












      • Ohhh, I see. $m$ is the gradient and I just need to multiply $|x|$ and $|y|$ by the components of the directional vector of $vec{BA}$. Thank you.
        – Lukas Knudsen
        Nov 15 at 17:18
















      Thank you so much for the answer. Is your labeling relative to the picture i posted? The math doesn't quite make sense to me. Sorry if it is too much to ask for.
      – Lukas Knudsen
      Nov 15 at 16:20




      Thank you so much for the answer. Is your labeling relative to the picture i posted? The math doesn't quite make sense to me. Sorry if it is too much to ask for.
      – Lukas Knudsen
      Nov 15 at 16:20












      @LukasKnudsen: You have not labeled vertices in your picture, I guess you want vector to start at $C$ according to the $angle{C}$ so just replace $B$ with $C$ in my answer. The other vertex remains $A$. The math is basically finding sine and cosine based on tangent.
      – Vasya
      Nov 15 at 16:32




      @LukasKnudsen: You have not labeled vertices in your picture, I guess you want vector to start at $C$ according to the $angle{C}$ so just replace $B$ with $C$ in my answer. The other vertex remains $A$. The math is basically finding sine and cosine based on tangent.
      – Vasya
      Nov 15 at 16:32












      Thank you for the clarification. Just to make sure I understood it, here's an example. $A( 0, 0 ), B( 10, 5 ), v=5, m=tan^-1(frac{5 - 0}{10 - 0})=26.6, |x|=frac{5}{sqrt{1 + 26.6^2}}=0.188, |y|=frac{26.6 * 5}{sqrt{1 + 26.6^2}}=4.996$. The x and y components seem to be off. Is this correct?
      – Lukas Knudsen
      Nov 15 at 16:58






      Thank you for the clarification. Just to make sure I understood it, here's an example. $A( 0, 0 ), B( 10, 5 ), v=5, m=tan^-1(frac{5 - 0}{10 - 0})=26.6, |x|=frac{5}{sqrt{1 + 26.6^2}}=0.188, |y|=frac{26.6 * 5}{sqrt{1 + 26.6^2}}=4.996$. The x and y components seem to be off. Is this correct?
      – Lukas Knudsen
      Nov 15 at 16:58














      @LukasKnudsen: no, $m=frac{5-0}{10-0}=0.5$, $|x|approx 4.47$, $|y| approx 2.236$, $BA(-10,-5)$ so $x=-4.47, y=-2.236$
      – Vasya
      Nov 15 at 17:12






      @LukasKnudsen: no, $m=frac{5-0}{10-0}=0.5$, $|x|approx 4.47$, $|y| approx 2.236$, $BA(-10,-5)$ so $x=-4.47, y=-2.236$
      – Vasya
      Nov 15 at 17:12














      Ohhh, I see. $m$ is the gradient and I just need to multiply $|x|$ and $|y|$ by the components of the directional vector of $vec{BA}$. Thank you.
      – Lukas Knudsen
      Nov 15 at 17:18




      Ohhh, I see. $m$ is the gradient and I just need to multiply $|x|$ and $|y|$ by the components of the directional vector of $vec{BA}$. Thank you.
      – Lukas Knudsen
      Nov 15 at 17:18










      up vote
      0
      down vote













      Not from the information provided. In particular, any rotation of the triangle shown would give different $x$ and $y$ values, but the same $A$, $B$, and $C$, so there is no possible function giving $x$ and $y$ in terms of $A$, $B$, and $C$ alone.






      share|cite|improve this answer

























        up vote
        0
        down vote













        Not from the information provided. In particular, any rotation of the triangle shown would give different $x$ and $y$ values, but the same $A$, $B$, and $C$, so there is no possible function giving $x$ and $y$ in terms of $A$, $B$, and $C$ alone.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          Not from the information provided. In particular, any rotation of the triangle shown would give different $x$ and $y$ values, but the same $A$, $B$, and $C$, so there is no possible function giving $x$ and $y$ in terms of $A$, $B$, and $C$ alone.






          share|cite|improve this answer












          Not from the information provided. In particular, any rotation of the triangle shown would give different $x$ and $y$ values, but the same $A$, $B$, and $C$, so there is no possible function giving $x$ and $y$ in terms of $A$, $B$, and $C$ alone.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 15 at 15:40









          user3482749

          1,709411




          1,709411






















              up vote
              0
              down vote













              Since you know the coordinates of the vertices, you can ignore the angles. The coordinates of $vec v$ can be found as a proportion of the coordinates of the appropriate edge. For the illustrated case, $$vec v = {|vec v|over|A-C|}(A-C).$$






              share|cite|improve this answer

























                up vote
                0
                down vote













                Since you know the coordinates of the vertices, you can ignore the angles. The coordinates of $vec v$ can be found as a proportion of the coordinates of the appropriate edge. For the illustrated case, $$vec v = {|vec v|over|A-C|}(A-C).$$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Since you know the coordinates of the vertices, you can ignore the angles. The coordinates of $vec v$ can be found as a proportion of the coordinates of the appropriate edge. For the illustrated case, $$vec v = {|vec v|over|A-C|}(A-C).$$






                  share|cite|improve this answer












                  Since you know the coordinates of the vertices, you can ignore the angles. The coordinates of $vec v$ can be found as a proportion of the coordinates of the appropriate edge. For the illustrated case, $$vec v = {|vec v|over|A-C|}(A-C).$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 16 at 2:39









                  amd

                  28.5k21049




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